representation on number line

(\le -4) gives a closed endpoint, and (>5) gives an open endpoint. In exams, always use round brackets on the infinite side.

The direct answer is option A: x<2, with an open point at 2 and shading to the left. Solve 7-3x>1. Subtract 7 from both sides: -3x>-6. Divide by -3; division by a negative number reverses the inequality, giving x<2. Since 2 itself is not allowed, the point at 2 is open. Values smaller than 2, such as 0, satisfy the inequality, so the shading goes left. Option A shows exactly this. Option B has the wrong direction because it says x>2. Option C includes 2, although strict > does not permit equality. Option D has both the wrong direction and an included endpoint. The key rule is that multiplying or dividing an inequality by a negative number reverses its sign.

The common part is from (2) to (3), but (2) is excluded and (3) is included. In exams, check endpoint inclusion separately in intersections.

A closed ray includes the endpoint, and a left ray means less than or equal. In exams, check both ray direction and endpoint.

Multiplying by (-3) reverses the inequality to (2x+5>-3), so (x>-4). In exams, reverse the sign when removing a negative denominator.

In (x<0), (0) is open; in (2\le x<5), (2) is closed and (5) is open. In exams, convert or into union.

Usually in Class (11), (\mathbb{N}={1,2,3,\ldots}), so points from (1) to (6) are marked. In exams, if the domain is natural numbers, mark only natural points.

The direct answer is option A: [4,10). Begin with 0≤(x-4)/3<2. Multiply every part by positive 3; the signs remain unchanged: 0≤x-4<6. Add 4 to all parts: 4≤x<10. Thus 4 is included and receives a closed point or square bracket, while 10 is excluded and receives an open point or round bracket. Option A is exactly [4,10). Option B incorrectly excludes 4 and includes 10, reversing both endpoint rules. Option C gives the bounds from the unsolved middle expression rather than the x-values. Option D represents values outside the interval and is unrelated to the solution set. A useful check is to test x=4: it gives 0, so it must be included; x=10 gives 2, but the upper sign is strict, so it must be excluded.

The two intervals meet at (-2), and (5) is not included. In exams, merge touching intervals when the common endpoint is included.

(x<1) and (x\ge 1) together give all real numbers. In exams, the union of complementary rays is the whole number line.

No (x) can be greater than (3) and less than or equal to (3) at the same time. In exams, the intersection of contradictory conditions is empty.

(2-2x<x+8) gives (-3x<6), so (x>-2). In exams, change the sign while dividing by a negative coefficient.

The common part is from (-5) to (1), including (-5) and excluding (1). In exams, both conditions must be satisfied in an intersection.

Open dots exclude the endpoints, and outer shading gives two rays. In exams, read outer shading as an or inequality.

The intervals overlap and together extend from (-4) to (3). In exams, the union of overlapping intervals forms one larger interval.

Removing ([-1,2)) removes (-1) but keeps (2). In exams, check endpoints carefully in set difference.

The points are (-3,-2,-1,0,1,2,3), so there are (7) points. In exams, include both endpoints while counting integers.

(4\le -2x<10) gives (-5<x\le -2) after negative division. In exams, endpoints also reverse when dividing a compound inequality by a negative.

The intervals join at (6) with no gap. In exams, intervals merge when the common endpoint is included.

The common part is from (2) to (5), including (2) and excluding (5). In exams, read intersection as a simultaneous condition.

Multiplying by (15) gives (3x+6\ge 5x-20), so (x\le 13), not (x\ge 13). In exams, recheck the final sign and arithmetic.

Multiplying by (4) gives (3x-1<2x+10), so (x<11). In exams, after clearing denominators, a strict sign gives an open endpoint.

Multiplying all parts by (2) gives (14\le 3x+2<32), so (4\le x<10). In exams, apply the same operation to all parts of a compound inequality.

Integers greater than (-\frac{9}{2}) start at (-4), and (2) is the last integer up to (\frac{5}{2}). In exams, choose the nearest valid integer at fractional boundaries.

The common part of the two intervals is from (-3) to (1), excluding (-3) and including (1). In exams, check endpoint inclusion separately in intersections.